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\begin{document}

\title{高等代数二}
\subtitle{8-3-正交变换 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月9日} }

\maketitle

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.3.i. 作业：星期天晚上十点半之前在网络教学平台提交 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}
\item   整理课堂笔记，补充没写完的计算或证明。
\item   习题(8.3)\#1,2,3,5,8, 抄写题目。
\end{enumerate}

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\begin{frame}{8.3.ii. 目录 }

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\begin{enumerate}

\item[8.3.1.] 正交变换的定义
\item[8.3.2.] 正交变换的例子
\item[8.3.6.] 定理8.3.1. 正交变换的充要条件
\item[8.3.8.] 定理8.3.2. 正交变换的充要条件
\item[8.3.9.] 定理8.3.3. 正交变换的充要条件

\end{enumerate}

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\begin{frame}{8.3.iii. 课堂讲解重点 }

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\begin{enumerate}
\item  正交变换的概念和例子
\item  正交变换的几个充分必要条件
\end{enumerate}

\end{frame}

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%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{8.3.1. 正交变换的定义}

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\begin{itemize}

\item  {\color{red}问题：一个 欧氏空间 $V$ 上的线性变换 $\sigma$ 什么时候称为正交变换？}

\item  解答：一个线性变换 $\sigma:V\to V$ 称为是正交变换，如果对任意向量 $\alpha\in V$, 线性变换 $\sigma$ 都保持 $\alpha$ 的长度不变，即 $$\lvert\sigma(\alpha)\rvert =\lvert \alpha \rvert. $$

\item  注：向量 $\alpha$ 的长度定义为 $$\lvert \alpha \rvert = \sqrt{\langle \alpha, \alpha \rangle}. $$ 

\item  要证明线性变换 $\sigma$ 是正交变换，就是要验证对任意 $\alpha\in V$, 都成立
$$\langle \sigma(\alpha), \sigma(\alpha) \rangle = \langle \alpha, \alpha \rangle. $$

\end{itemize}

\end{frame}

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\begin{frame}{8.3.2. 例子 }

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\begin{itemize}

\item  {\color{red}问题：考虑欧氏空间 $\mathbb{R}^2$ 到自身的线性变换 
$$\sigma: (x,y)\mapsto (x\cos\theta-y\sin\theta, x\sin\theta+y\cos\theta),$$
证明这是一个正交变换。
}

\item  解答：按照定义直接验证。
\begin{eqnarray*}
v &=& (x,y), \\
\sigma(v) &=& (x\cos\theta-y\sin\theta, x\sin\theta+y\cos\theta), \\ 
\lvert \sigma(v)\rvert ^2 &=& (x\cos\theta-y\sin\theta)^2 + (x\sin\theta+y\cos\theta)^2 \\ 
&=& x^2+y^2 \\
&=& \lvert v \rvert ^2. 
\end{eqnarray*}

\end{itemize}

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\begin{frame}{8.3.3. 例子 }

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\begin{itemize}

\item  {\color{red}问题：
考虑欧氏空间 $V=\mathbb{R}^3$. 设 $W$ 是由平面 $z=0$ 所定义。设变换 $\sigma:V\to V$ 是相对于这个平面的镜面反射。}
\begin{enumerate}
\item {\color{red}设 $\alpha=(x,y,z)$, 写出 $\sigma(\alpha)$ 的具体表达式。}
\item {\color{red}验证这个变换是一个正交变换。}
\end{enumerate}

\item  解答：
\begin{enumerate}
\item  $\sigma((x,y,z)) = (x,y,-z)$. 
\item  $\lvert \sigma((x,y,z)) \rvert ^2 = x^2+y^2+z^2 = \lvert (x,y,z) \rvert ^2$. 
\end{enumerate}

\vfill 

\item  注：将平面 $z=0$ 换成平面 $x+y+z=0$, 情况又如何？

\end{itemize}

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\begin{frame}{8.3.4. 例子 }

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\begin{itemize}
\item  {\color{red}问题：设 $\sigma:\mathbb{R}^2 \to \mathbb{R}^2$ 是关于直线 $y=0$ 的反射。写出这个正交变换在标准基 $\{\varepsilon_1=(1,0),\varepsilon_2=(0,1)\}$ 下的矩阵。}

\item  解答：关于直线 $y=0$ 的反射可以写成 $\sigma((x,y)) = (-x,y)$. 因此 
{\footnotesize 
\begin{eqnarray*}
(\sigma(\varepsilon_1), \sigma(\varepsilon_2) ) 
&=& (\sigma((1,0)), \sigma((0,1)) )  \\ 
&=& ((-1,0), (0,1))  \\ 
&=& (\varepsilon_1, \varepsilon_2)
\begin{pmatrix} -1&0 \\ 0&1 \end{pmatrix}. 
\end{eqnarray*}
}

\vspace{-0.6cm}

%\begin{center}
\includegraphics[height=0.4\textheight,width=0.4\textwidth]{pic/orth-transform-0.png}
%\end{center}

\end{itemize}

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\begin{frame}{8.3.5. 例子 }

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\begin{itemize}
\item  {\color{red}问题：设 $\sigma:\mathbb{R}^3 \to \mathbb{R}^3$ 是绕直线 $x=y=0$ 的按右手法则的旋转 $\theta$ 角度。写出这个正交变换关于标准基 $$\{\varepsilon_1=(1,0,0),\varepsilon_2=(0,1,0),\varepsilon_3=(0,0,1)\}$$ 的矩阵。
}

\item  解答：坐标轴 $Oz$ 轴是保持不动的。坐标平面 $Oxy$ 平面也是不变子空间。
{\footnotesize 
\begin{eqnarray*}
(\sigma(\varepsilon_1), \sigma(\varepsilon_2), \sigma(\varepsilon_3) ) 
&=& (\varepsilon_1, \varepsilon_2, \varepsilon_3)
\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta&\cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}


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%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{8.3.6. 定理8.3.1. 正交变换的充要条件 }

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\begin{itemize}

\item  {\color{red}定理：
设 $\sigma:V\to V$ 是欧氏空间到自身的线性变换。则下述条件等价：}
\begin{enumerate}[(a)]
\item {\color{red}这个线性变换是一个正交变换。}
\item {\color{red}对任意的向量 $\alpha,\beta\in V$, 都有 $\langle\sigma(\alpha),\sigma(\beta)\rangle = \langle\alpha,\beta\rangle$. }
\end{enumerate}

\item  证明：
\begin{enumerate}
\item (b) $\Rightarrow$ (a): 取 $\beta = \alpha$, 即得。
\item (a) $\Rightarrow$ (b): 由恒等式 $4\langle \xi, \eta \rangle = \lvert \xi+\eta \rvert ^2 - \lvert \xi-\eta \rvert ^2$ 可得
\begin{eqnarray*}
4 \langle\sigma(\alpha),\sigma(\beta)\rangle 
&=& \lvert \sigma(\alpha)+\sigma(\beta) \rvert ^2 - \lvert \sigma(\alpha)-\sigma(\beta) \rvert ^2 \\ 
&=& \lvert \sigma(\alpha+\beta) \rvert ^2 - \lvert \sigma(\alpha-\beta) \rvert ^2 \\ 
&=& \lvert \alpha+\beta \rvert ^2 - \lvert \alpha-\beta \rvert ^2 \\ 
&=& 4  \langle\alpha,\beta\rangle.  
\end{eqnarray*}

\end{enumerate}


\end{itemize}

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%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{8.3.7. 一个记号 }

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\begin{itemize}

\item  {\color{red}问题：设欧氏空间中有两个向量组 
\begin{align*}
\Phi = \{\alpha_1, \alpha_2, \cdots, \alpha_n\}, \hspace{0.3cm} 
\Psi = \{\beta_1, \beta_2, \cdots, \beta_n\}. 
\end{align*} 
记号 $\langle \Phi^t, \Psi \rangle$ 的含义是什么？}

\item  解答：这是由第 $(i,j)$ 元素为内积 $\langle \alpha_i, \beta_j \rangle $ 构成的矩阵，即 
{\footnotesize 
\begin{align*}
\langle \Phi^t, \Psi \rangle := \begin{pmatrix} 
\langle \alpha_1, \beta_1 \rangle & \langle \alpha_1, \beta_2 \rangle & \cdots & \langle \alpha_1, \beta_n \rangle \\ 
\langle \alpha_2, \beta_1 \rangle & \langle \alpha_2, \beta_2 \rangle & \cdots & \langle \alpha_2, \beta_n \rangle \\ 
\vdots & \vdots & & \vdots & \\ 
\langle \alpha_n, \beta_1 \rangle & \langle \alpha_n, \beta_2 \rangle & \cdots & \langle \alpha_n, \beta_n \rangle \\ 
\end{pmatrix}. 
\end{align*} 
}

\end{itemize}

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%\begin{frame}[fragile=singleslide]{6.1.1. }
\begin{frame}{8.3.8. 定理8.3.2. 正交变换的另一个充要条件 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}定理：
设 $\sigma:V\to V$ 是欧氏空间到自身的线性变换。则下述条件等价：}
\begin{enumerate}[(a)]
\item  {\color{red}这个线性变换是一个正交变换。}
\item  {\color{red}这个线性变换把任意规范正交基映成规范正交基。}
\item  {\color{red}这个线性变换把某个规范正交基映成规范正交基。}
\end{enumerate}

\item  证明：
\begin{enumerate}
\item (a) $\Rightarrow$ (b): 
\begin{enumerate}
\item  设 $\Phi$ 是任意一个规范正交基，则 $\langle \Phi^t, \Phi \rangle = E$. 
\item  因为 $\sigma$ 是正交变换，所以 $\langle \sigma(\Phi)^t, \sigma(\Phi) \rangle =\langle \Phi^t, \Phi \rangle$.
\item  所以 $\langle \sigma(\Phi)^t, \sigma(\Phi) \rangle = E$, 即 $\sigma(\Phi)$ 是规范正交基。
\end{enumerate}
%\item (b) $\Rightarrow$ (c): 基本逻辑。
\item (c) $\Rightarrow$ (a): 
\begin{enumerate}
\item  设 $\Phi$ 是某个规范正交基，且 $\sigma(\Phi)$ 仍是规范正交基。
\item  则 $\langle \sigma(\Phi)^t, \sigma(\Phi) \rangle = E$, $\langle \Phi^t, \Phi \rangle = E$.
\item  对任意 $\alpha\in V$, 设 $\alpha=\Phi\cdot X$, 则有 

\vspace{-0.6cm}
$$\langle\sigma(\alpha),\sigma(\alpha)\rangle 
= \langle X^t\sigma(\Phi)^t, \sigma(\Phi)X \rangle 
= X^t\langle \sigma(\Phi)^t, \sigma(\Phi) \rangle X 
= X^tEX 
= \langle\alpha,\alpha\rangle. $$
\end{enumerate}

\end{enumerate}


\end{itemize}

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\begin{frame}{8.3.9. 定理8.3.3. 正交变换的又一个充要条件 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}定理：
设 $\sigma:V\to V$ 是欧氏空间到自身的线性变换。则下述条件等价：}
\begin{enumerate}[(a)]
\item  {\color{red}这个线性变换是一个正交变换。}
\item  {\color{red}这个线性变换在任意规范正交基下的矩阵是正交矩阵。}
\item  {\color{red}这个线性变换在某个规范正交基下的矩阵是正交矩阵。}
\end{enumerate}

\item  证明：
\begin{enumerate}
\item (a) $\Rightarrow$ (b): 
\begin{enumerate}
\item  设 $\Phi$ 是任意一个规范正交基，设 $\sigma(\Phi) = \Phi\cdot A$. 
\item  因为 $\sigma$ 是正交变换，所以 $\langle \sigma(\Phi)^t, \sigma(\Phi) \rangle =\langle \Phi^t, \Phi \rangle = E$.
\item  所以 $A^tA = E$, 即 $A$ 是正交矩阵。
\end{enumerate}
%\item (b) $\Rightarrow$ (c): 基本逻辑。
\item (c) $\Rightarrow$ (a): 
\begin{enumerate}
\item  设 $\Phi$ 是某个规范正交基，且 $\sigma(\Phi)=\Phi\cdot A$, 其中 $A$ 是正交矩阵。
\item  则 $\langle \sigma(\Phi)^t, \sigma(\Phi) \rangle = A^t \langle \Phi^t, \Phi \rangle A = A^tEA=A^tA=E$.
\item  因此 $\sigma(\Phi)$ 也是一个规范正交基。
%\item  可证对任意 $\alpha\in V$, 有 $\langle\sigma(\alpha),\sigma(\alpha)\rangle = \langle\alpha,\alpha\rangle$. 
\end{enumerate}

\end{enumerate}


\end{itemize}

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\begin{frame}{习题(8.3)\#1 }

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\begin{itemize}

\item  {\color{red}问题：设 $\sigma, \tau : V\to V$ 是两个正交变换。} 
\begin{enumerate}
\item  {\color{red}证明乘积 $\sigma\circ \tau$ 仍是正交变换。}
\item  {\color{red}证明逆变换 $\sigma^{-1}$ 仍是正交变换。}
\end{enumerate}

\item  思路：验证正交变换的定义。或验证正交变换的充分必要条件。


\end{itemize}

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\begin{frame}{习题(8.3)\#2 }

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\begin{itemize}

\item  {\color{red}问题：设 $\sigma$ 是 $n$ 维欧氏空间 $V$的一个正交变换。设 $W$ 是 $\sigma$ 的不变子空间，证明正交补空间 $W{\,}^\perp$ 也是 $\sigma$ 的不变子空间。

}

\item  思路：根据正交补空间的定义，和不变子空间的定义。


\end{itemize}

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\begin{frame}{习题(8.3)\#3 }

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\begin{itemize}

\item  {\color{red}问题：设 $V$ 是一个欧氏空间，设 $\alpha\in V$ 是一个非零向量。
定义 $$\tau(\xi)=\xi - \frac{2\langle \xi,\alpha \rangle }{\langle \alpha,\alpha \rangle }\alpha. $$
证明 $\tau$ 是一个正交变换，且 $\tau^2=\iota$ 是恒等变换。
}

\item  思路：直接验证正交变换的定义。为简化计算，可设 $\alpha=k\beta$, 其中 $\beta$ 是单位向量。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $A$ 是一个三阶正交矩阵，且 $\det(A)=1$. 
\begin{enumerate}
\item  {\color{red}证明 $A$ 有一个特征值等于1. } 
\item  {\color{red}证明 $A$ 的特征多项式有形状 $f(x)=x^3-tx^2+tx-1$, 其中 $-1\le t\le 3$. } 
\end{enumerate}

}

\item  思路：选取 $V=\mathbb{R}^3$ 的一个适当的基，使得 $P^{-1}AP$ 为实的若而当标准形。


\end{itemize}

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\begin{frame}{习题(8.3)\#8 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：证明每一个 $n$ 阶非奇异实数矩阵 $A$ 都可以唯一地表示成 $A=UT$ 的形式，其中 $U$ 是一个正交矩阵，$T$ 是一个上三角形实数矩阵，	且主对角线上的元素都是正数。
}

\item  思路：将斯密特正交化方法和规范化方法写成矩阵乘法。例如，
{\footnotesize 
\begin{eqnarray*}
\beta_1 &=& \alpha_1 \\ 
\beta_2 &=& \alpha_2- k_1\beta_1 \\ 
\beta_3 &=& \alpha_3- \ell_1\beta_1-\ell_2\beta_2 
\end{eqnarray*}
}
可以写成
{\footnotesize 
\begin{eqnarray*}
(\beta_1, \beta_2, \beta_3 ) = (\alpha_1, \alpha_2, \alpha_3) \begin{pmatrix} 1&*&* \\ 0&1 & * \\ 0&0&1 \end{pmatrix}. 
\end{eqnarray*}
}

\end{itemize}

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\begin{frame}{ Rules of Inference }

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\begin{center}
\begin{tabular}{|l|l|} \hline 
Name & Tautology \\ \hline 
Modus ponens & $p\wedge (p\to q) \to q$ \\ \hline 
Modus tollens & $\neg q \wedge (p\to q) \to \neg p$  \\ \hline 
Hypothetical syllogism & $((p\to q) \wedge (q\to r)) \to (p\to r)$ \\ \hline 
Disjunctive syllogism & $((p\vee q) \wedge \neg p) \to q$ \\ \hline 
Addition & $p \to (p\vee q)$ \\ \hline 
Simplification & $(p\wedge q) \to p$ \\ \hline 
Conjunction & $((p)\wedge (q)) \to (p\wedge q)$ \\ \hline 
Resolution & $((p\vee q) \wedge (\neg p \vee r)) \to (q\vee r)$ \\ \hline 
\end{tabular}

\end{center}


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